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Chapter 3 Geometry of convex functions - Meboo Publishing ...

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226 CHAPTER 3. GEOMETRY OF CONVEX FUNCTIONS<br />

3.3.2.1.3 Exercise. k-largest norm gradient.<br />

Prove (510). Find ∇‖x‖ 1 and ∇‖x‖n<br />

k<br />

on R n . 3.11<br />

<br />

3.3.3 clipping<br />

Zeroing negative vector entries under 1-norm is accomplished:<br />

‖x + ‖ 1 = minimize 1 T t<br />

t∈R n<br />

subject to x ≼ t<br />

0 ≼ t<br />

(512)<br />

where, for x=[x i , i=1... n]∈ R n<br />

x + t ⋆ =<br />

(clipping)<br />

[<br />

xi , x i ≥ 0<br />

0, x i < 0<br />

}<br />

, i=1... n<br />

]<br />

= 1 (x + |x|) (513)<br />

2<br />

minimize<br />

x∈R n ‖x + ‖ 1<br />

subject to x ∈ C<br />

≡<br />

minimize 1 T t<br />

x∈R n , t∈R n<br />

subject to x ≼ t<br />

0 ≼ t<br />

x ∈ C<br />

(514)<br />

3.4 Inverted <strong>functions</strong> and roots<br />

A given function f is <strong>convex</strong> iff −f is concave. Both <strong>functions</strong> are loosely<br />

referred to as <strong>convex</strong> since −f is simply f inverted about the abscissa axis,<br />

and minimization <strong>of</strong> f is equivalent to maximization <strong>of</strong> −f .<br />

A given positive function f is <strong>convex</strong> iff 1/f is concave; f inverted about<br />

ordinate 1 is concave. Minimization <strong>of</strong> f is maximization <strong>of</strong> 1/f .<br />

We wish to implement objectives <strong>of</strong> the form x −1 . Suppose we have a<br />

2×2 matrix<br />

[ ] x z<br />

T ∈ R 2 (515)<br />

z y<br />

3.11 Hint:D.2.1.

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